Question: $ \left(\dfrac{16}{81}\right)^{-\frac{3}{4}}$
Solution: $= \left(\dfrac{81}{16}\right)^{\frac{3}{4}}$ $= \left(\left(\dfrac{81}{16}\right)^{\frac{1}{4}}\right)^{3}$ To simplify $\left(\dfrac{81}{16}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left(? \right)^{4}=\dfrac{81}{16}$ To simplify $\left(\dfrac{81}{16}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left({\dfrac{3}{2}}\right)^{4}=\dfrac{81}{16}$ so $ \left(\dfrac{81}{16}\right)^{\frac{1}{4}}=\dfrac{3}{2}$ So $\left(\dfrac{81}{16}\right)^{\frac{3}{4}}=\left(\left(\dfrac{81}{16}\right)^{\frac{1}{4}}\right)^{3}=\left(\dfrac{3}{2}\right)^{3}$ $= \left(\dfrac{3}{2}\right)\cdot\left(\dfrac{3}{2}\right)\cdot \left(\dfrac{3}{2}\right)$ $= \dfrac{9}{4}\cdot\left(\dfrac{3}{2}\right)$ $= \dfrac{27}{8}$